Page 387 - Physics Coursebook 2015 (A level)

Chapter 24: Capacitance
QUESTIONS
1 Calculate the charge on a 220 μF capacitor charged up to 15 V. Give your answer in microcoulombs (μC) and in coulombs (C).
2 A charge of 1.0 × 10−3 C is measured on a capacitor with a potential difference across it of 500 V. Calculate the capacitance in farads (F), microfarads (μF) and picofarads (pF).
3 Calculate the average current required to charge a 50 μF capacitor to a p.d. of 10 V in a time interval of0.01s.
4 A student connects an uncharged capacitor of capacitance C in series with a resistor, a cell and a switch. The student closes the switch and records the current I at intervals of 10 s. The results are shown in Table 24.1. The potential difference across the capacitor after 60 s was 8.5 V. Plot a current–time graph, and use it to estimate the value of C.
t/s I/μA
Table 24.1 Data for Question 4.
Energy stored in a capacitor
When you charge a capacitor, you use a power supply to push electrons onto one plate and off the other. The power supply does work on the electrons, so their potential energy increases. You recover this energy when you discharge the capacitor.
If you charge a large capacitor (1000 μF or more) to a potential difference of 6.0 V, disconnect it from the supply, and then connect it across a 6.0 V lamp, you can see the lamp glow as energy is released from the capacitor. The lamp will flash briefly. Clearly, such a capacitor does not store much energy when it is charged.
In order to charge a capacitor, work must be done to push electrons onto one plate and off the other (Figure 24.6). At first, there is only a small amount of negative charge on the left-hand plate. Adding more electrons is relatively easy, because there is not much repulsion. As
the charge on the plate increases, the repulsion between the electrons on the plate and the new electrons increases, and a greater amount of work must be done to increase the charge on the plate.
This can be seen qualitatively in Figure 24.7a. This graph shows how the p.d. V increases as the amount of
other electrons
on plate repel – force this electron –
pushing –
electrons –
– +
Figure 24.6 When a capacitor is charged, work must be done to push additional electrons against the repulsion of the electrons that are already present.
charge Q increases. It is a straight line because Q and V are related by:
V = QC
We can use Figure 24.7a to calculate the work done in charging up the capacitor.
First, consider the work done W in moving charge Q through a constant p.d. V. This is given by:
W = QV
(You studied this equation in Chapter 9.) From the graph of Q against V (Figure 24.7b), we can see that the quantity Q × V is given by the area under the graph.
+
+ force
– pulling
+ electrons
0
10
20
30
40
50
60
200
142
102
75
51
37
27
The area under a graph of p.d. against charge is equal to work done.
aV
energy stored
00
bV
0
Q
energy required to move charge Q through p.d. V
0Q
Figure 24.7 The area under a graph of voltage against charge gives a quantity of energy. The area in a shows the energy stored in a capacitor; the area in b shows the energy required to drive a charge through a resistor.
375